Problem: What is the value of $99^3 + 3(99^2) + 3(99) + 1$?
The given expression is the expansion of $(99+1)^3$.  In general, the cube $(x+y)^3$ is \[(x+y)^3=1x^3+3x^2y+3xy^2+1y^3.\]   The first and last terms in the given expression are cubes and the middle two terms both have coefficient 3, giving us a clue that this is a cube of a binomial and can be written in the form \[(x+y)^3\]In this case, $x=99$ and $y=1$, so our answer is\[(99+1)^3\ = 100^3 = \boxed{1,\!000,\!000}\]